Subsequences: Picking in Order

Pick terms out of a sequence while keeping their order, and a new sequence appears. That is a subsequence.

The indices only need to grow strictly. Terms may be skipped, but the order may never change. If the original sequence converges, every subsequence converges to the same value. Conversely, if two subsequences head to different values, the original diverges.

For example, an 🚃 express train does not stop at every station. Line up only the stations where it stops, in order, and you get a subsequence of the local line.

n1<n2<n3<⋯ ,(ank)k=1∞n_1 < n_2 < n_3 < \cdots, \qquad (a_{n_k})_{k=1}^{\infty}
an→L ⟹ ank→La_n \to L \ \Longrightarrow \ a_{n_k} \to L

Examples & Exercises

Prove completely that if an→La_n \to L, then every subsequence anka_{n_k} also converges to LL.

an→L ⟹ ank→La_n \to L \ \Longrightarrow \ a_{n_k} \to L
The picked even terms still head to their crowd's limit.
Show proof

  1. First, a fact about indices: n1<n2<⋯n_1 < n_2 < \cdots strictly increases in the naturals, so nk≥kn_k \ge k for every kk. By induction on kk: n1≥1n_1 \ge 1, and nk+1≥nk+1≥k+1n_{k + 1} \ge n_k + 1 \ge k + 1.

  2. Given ε>0\varepsilon > 0, convergence an→La_n \to L gives NN with ∣an−L∣<ε|a_n - L| < \varepsilon for n>Nn > N. For k>Nk > N we have nk≥k>Nn_k \ge k > N, so ∣ank−L∣<ε|a_{n_k} - L| < \varepsilon.

  3. The same NN works for the subsequence. Hence ank→La_{n_k} \to L.

Find the limits of the even and odd subsequences of an=(−1)n+1na_n = (-1)^n + \frac{1}{n}, and prove completely that the original sequence diverges.

a2k=1+12k→1,a2k−1=−1+12k−1→−1a_{2k} = 1 + \frac{1}{2k} \to 1, \qquad a_{2k - 1} = -1 + \frac{1}{2k - 1} \to -1
The full picture: the two crowds split apart.
Show proof

  1. Even terms: a2k=1+12ka_{2k} = 1 + \frac{1}{2k} with 12k→0\frac{1}{2k} \to 0, so a2k→1a_{2k} \to 1. Odd terms: a2k−1=−1+12k−1a_{2k - 1} = -1 + \frac{1}{2k - 1} with 12k−1→0\frac{1}{2k - 1} \to 0, so a2k−1→−1a_{2k - 1} \to -1.

  2. If the original sequence converged to some LL, the previous problem forces every subsequence to converge to LL. But the even subsequence tends to 1 and the odd one to −1-1. Since 1≠−11 \ne -1, they cannot both equal LL.

  3. The contradiction shows the original sequence converges to nothing. It diverges.

Prove completely that if a2k→La_{2k} \to L and a2k−1→La_{2k - 1} \to L, then an→La_n \to L.

a2k→L, a2k−1→L ⟹ an→La_{2k} \to L, \ a_{2k - 1} \to L \ \Longrightarrow \ a_n \to L
When the odd terms head to the same value, the whole sequence follows.
Show proof

  1. Given ε>0\varepsilon > 0, convergence of the even subsequence gives K1K_1 with ∣a2k−L∣<ε|a_{2k} - L| < \varepsilon for k>K1k > K_1, and convergence of the odd subsequence gives K2K_2 with ∣a2k−1−L∣<ε|a_{2k - 1} - L| < \varepsilon for k>K2k > K_2.

  2. Put N=2max⁡(K1,K2)N = 2 \max(K_1, K_2). Any n>Nn > N is even or odd. If n=2kn = 2k then k>K1k > K_1 and the bound holds; if n=2k−1n = 2k - 1 then k>K2k > K_2 and it holds. Either way ∣an−L∣<ε|a_n - L| < \varepsilon.

  3. Since ε\varepsilon was arbitrary, an→La_n \to L.

3. Sequences